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Question
An observed from the top of a 150 m tall lighthouse, the angles of depression of two ships approaching it are 30° and 45°. If one ship is directly behind the other, find the distance between the two ships.
Solution
Let AB be the lighthouse of 150 m. and angle of depression of two ship C and D are 30° and 45° respectively.
let BC = x,CD = y and ∠ADB = 30°, ∠ACB = 45°
We use trigonometric ratios.
IN a triangle ABC
`=> tan 45^@ = (AB)/(BC)`
`=> 1 = 150/x`
`=> x = 150`
Again in a triangle ABD
`=> tan 30° = (AB)/(BD)`
`=> 1/sqrt3 = 150/(x + y)`
`=> x + y = 150sqrt3`
`=> 150 + y = 150sqrt3`
`=> y = 150sqrt3 - 150`
`=> y = 150(sqrt3 - 1)`
`=> y = 150 xx 0.732`
Hence distance between the ships is 109.8 m
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